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Rope hanging from a table, find time taken to fall off table

  1. Sep 14, 2015 #1
    1. The problem statement, all variables and given/known data

    A rope of uniform mass per unit length ##\lambda## and length ##L## lies curled on a table, with a short length ##\delta l## of it hanging through a hole in the table. At time zero, the rope begins to slide through the hole. At what time ##t_c## (in seconds) does the end of the rope pass through the hole?

    Details




      • The rope is 10 m long.
      • ##\delta l = 0.1## m
      • ##g=9.81\ \text{m/s}^2##
      • There is no friction between the rope and itself, or between the rope and the table.


    2. Relevant equations


    3. The attempt at a solution

    I tried the problem but my solution doesn't give the correct answer. I would be truly grateful if somebody could please show me where I've gone wrong.


    Let the 0 Gravitational Potential Energy level be at the table's height.
    Taking the rope as the system on a table, we can see that only gravity does work on the system.
    Let ##x## be the part of the rope hanging through the hole. Consider an elemental part ##dm## of width ##dx## at a distance ##x## hanging below the table. $$dm=\lambda dx$$
    $$dW_g=xdm=\lambda g x dx$$
    Since ##x## varies from ##\delta l## to ##l## ie ##0.1## to ##10,##
    $$W_g=\lambda g\int_{0.1}^{10} xdx$$
    $$W_g=49.995\lambda g$$
    Applying the work Energy Theorem,
    $$\dfrac{1}{2}mv^2=W_g$$
    $$\Rightarrow \dfrac{1}{2}\times(\lambda\times L)\times v^2=49.995\lambda g$$
    Since $$\lambda$$ gets cancelled on both sides and ##L=10,##
    $$v^2=\dfrac{49.995}{5}\times g$$
    Hence, $$v=\sqrt{98.09}\approx9.904$$
    Since ##v=\dfrac{dx}{dt}##
    $$\dfrac{dx}{9.904}=dt$$
    $$\dfrac{1}{9.904}\int_{0.1}^{10}dx=\int _0^Tdt$$

    However, this answer doesn't match the given answer of 5.29.
     
  2. jcsd
  3. Sep 14, 2015 #2

    BvU

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    Hi BW,

    Your post reminded me of this one which isn't exactly the same, but perhaps you can pick up something useful. In the mean time I'll try to follow through your solution.
     
    Last edited: Sep 14, 2015
  4. Sep 14, 2015 #3

    andrewkirk

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    I think your work integral needs to be a double integral, not a single, because each rope element only generates a gravitational force when it is through the hole, but every element is subject to the force, via the tension in the rope, at all times. The double integral would be over (1) location of the element on the rope and (2) distance travelled by the element. However, I think that way of doing it may be unnecessarily complex.

    Can you write a differential equation expressing the acceleration of the rope (##\ddot{x}##) as a function of how much of the rope is hanging through the hole (##x##)? Unless I've made a mistake, the solution is fairly quick and easy via that route.
     
    Last edited: Sep 14, 2015
  5. Sep 14, 2015 #4
    Sorry Sir, I haven't done Differential Equations.
    Sir, I don't think a double integral would be necessary. It s true that the mass of the rope hanging is continuously changing and so is ##x##. However, we could write the elemental mass as ##dm=\dfrac{M}{L}\times dx## where ##dx## is the length of the element of mass. Thus, the expression for work done (##dm\times g\times x=\dfrac{M}{L}\times dx\times g\times x##) is reduced to only one quantity which is varying ie ##x##-the length of the rope above the element of mass.
     
  6. Sep 14, 2015 #5
    Sorry Sir, but that thread didn't really help me.
     
  7. Sep 14, 2015 #6

    BvU

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    Fair enough. From your solution attempt I gather you use the kinetic energy (##=W_g##) at the moment when the end of the rope falls out of the hole to calculate a velocity. But that is not the average velocity.
     
  8. Sep 14, 2015 #7

    andrewkirk

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    The work done on what, by what, within what constraints? If we integrate it, to what physical quantity does the result correspond?

    When we write an equation involving increments like this, we need to be able to relate it to an intuitive physical situation. If we just integrate it without stopping to think what it means, and what integrating it means, we are likely to end up with a meaningless number.
     
  9. Sep 14, 2015 #8
    In that case Sir, how would we find the average velocity?
    Ps. Sir, please could you also help me with [this question](https://www.physicsforums.com/threa...of-energy-and-momentum.832277/#post-5227689)?
     
    Last edited: Sep 14, 2015
  10. Sep 14, 2015 #9
    You are right Sir. Sir, in this case, the work is done by gravity on the hanging part of the rope. Also, the velocity I got corresponds to the velocity of the rope at the time the entire rope slips off the table.
    Please could you explain what you meant by 'within what constraints'? I couldn't quite understand.
    Sir would it also be possible for you to kidlyhelp me with [this question](https://www.physicsforums.com/threa...of-energy-and-momentum.832277/#post-5227689)?
     
    Last edited: Sep 14, 2015
  11. Sep 14, 2015 #10

    andrewkirk

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    How do you know?
    To talk about work done we need to make clear what is doing the work, what it it is doing the work on, and over what period or in what locations the work is being added up (integrated) to get a total work amount. The constraints refers to the locations over which the the work is being integrated/summed.
     
  12. Sep 14, 2015 #11
    Sir since the limit ##l=10## corresponds to the time when the entire rope slips off the table.
     
  13. Sep 14, 2015 #12
    Hello ,

    Please note that energy is not conserved in this problem . You would be better off going with the momentum approach .
     
  14. Sep 15, 2015 #13

    haruspex

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    I agree. But it still leads to a differential equation.
    Better WOrld, you can solve the problem by (a) using conservation of momentum and (b) assuming the acceleration is constant.
    At any instant, the weight of the suspended part (##\lambda g x##) is doing two things. It is accelerating the part already suspended, and it is accelerating the next little bit of the rope from rest to the current speed. Can you write out expressions for those two components of the total force?
     
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